Question

Obtain the zeros of the quadratic polynomial
√3x² − 8x + 4
√3 and verify relation between its zeros and coefficients.​

Answer 1

S O L U T I O N :

We have quadratic polynomial p(x) = √3x² - 8x + 4√3 & zero of the polynomial p(x) = 0.

$\underline{\underline{\tt{Using\:by\:quadratic\:formula\::}}}$

$\boxed{\bf{x = \frac{-b\pm \sqrt{b^{2} - 4ac} }{2a} }}$

As we know that quadratic polynomial compared with ax² + bx + c;

• a = √3
• b = -8
• c = 4√3

Now,

$\mapsto\sf{x = \dfrac{-(-8) \pm \sqrt{(-8)^{2} - 4 \times \sqrt{3} \times 4 \sqrt{3} } }{2 \times \sqrt{3} } }$

$\mapsto\sf{x = \dfrac{8 \pm \sqrt{64 - 4 \times \sqrt{3} \times 4 \sqrt{3} } }{2 \times \sqrt{3} } }$

$\mapsto\sf{x = \dfrac{8 \pm \sqrt{64 - 16 \times 3} }{2 \times \sqrt{3} } }$

$\mapsto\sf{x = \dfrac{8 \pm \sqrt{64 - 48} }{2 \times \sqrt{3} } }$

$\mapsto\sf{x = \dfrac{8 \pm \sqrt{16} }{2 \times \sqrt{3} } }$

$\mapsto\sf{x = \dfrac{8 \pm 4 }{2 \sqrt{3} } }$

$\mapsto\sf{x = \dfrac{8 + 4 }{2 \sqrt{3} }\:\:\:Or\:\:\:x = \dfrac{8 - 4 }{2 \sqrt{3} }}$

$\mapsto\sf{x = \dfrac{12 }{2 \sqrt{3} }\:\:\:Or\:\:\:x = \dfrac{4 }{2 \sqrt{3} }}$

∴ α = 12/2√3 & β = 4/2√3 are the zeroes of the given polynomials .

V E R I F I C A T I O N :

$\underline{\mathcal{SUM\:OF\:THE\:ZEROES\::}}$

$\mapsto\tt{\alpha + \beta = \dfrac{-b}{a} =\bigg\lgroup\dfrac{Coefficient\:of\:x}{Coefficient\:of\:x^{2} }\bigg\rgroup }$

$\mapsto\tt{\dfrac{12}{2\sqrt{3} } + \dfrac{4}{2\sqrt{3} } = \dfrac{-(-8)}{\sqrt{3} } }$

$\mapsto\tt{\dfrac{12+4}{2\sqrt{3} } = \dfrac{8}{\sqrt{3} } }$

$\mapsto\tt{\dfrac{16}{2\sqrt{3} } = \dfrac{8}{\sqrt{3} } }$

$\mapsto\tt{\dfrac{\cancel{16}}{\cancel{2}\sqrt{3} } = \dfrac{8}{\sqrt{3} } }$

$\mapsto\bf{\dfrac{8}{\sqrt{3} } = \dfrac{8}{\sqrt{3} } }$

$\underline{\mathcal{PRODUCT\:OF\:THE\:ZEROES\::}}$

$\mapsto\tt{\alpha \times \beta = \dfrac{c}{a} =\bigg\lgroup\dfrac{Constant\:term}{Coefficient\:of\:x^{2} }\bigg\rgroup }$

$\mapsto\tt{\dfrac{12}{2\sqrt{3} } \times \dfrac{4}{2\sqrt{3} } = \dfrac{4\sqrt{3} }{\sqrt{3} } }$

$\mapsto\tt{\dfrac{48}{4 \times 3 } = \dfrac{4\sqrt{3} }{\sqrt{3} } }$

$\mapsto\tt{\cancel{\dfrac{48}{12 }} = \dfrac{4\cancel{\sqrt{3} }}{\cancel{\sqrt{3} }} }$

$\mapsto\bf{4 = 4}$

Thus,

The relationship between zeroes & coefficient are verified .